Four Checks Before Accepting a Homotopy
A formula can look like a deformation without defining a homotopy. It may fail to be continuous, end at the wrong map, move endpoints that are required to stay fixed, or take values outside the proposed target space. These are separate tests: passing three does not compensate for failing the fourth.
Throughout, \(I=[0,1]\) has its subspace topology from the standard topology on \(\mathbb{R}\). For topological spaces \(X\) and \(Y\), the product \(X\times I\) has the product topology.
For paths, the domain is \(I\). A path homotopy from \(\alpha\) to \(\beta\) with endpoints fixed must satisfy \(H(0,s)=\alpha(0)=\beta(0)\) and \(H(1,s)=\alpha(1)=\beta(1)\) for every \(s\in I\). Thus checking only \(H(t,0)\) and \(H(t,1)\) does not check whether the endpoints stay fixed at intermediate parameter values.
Proof. The continuity and endpoint equalities are part of the definition of a homotopy. If \(H\) is relative to \(A\), its defining condition gives \(H(a,s)=f(a)=g(a)\) for every \(a\in A\) and \(s\in I\). Taking \(s=1\) gives \(g(a)=f(a)\), so \(f|_A=g|_A\); the same condition gives the required fixed values at every intermediate parameter. \(\square\)
This theorem is a checklist, not a way to infer that a formula works from its appearance. Each condition must be verified for the particular formula and the particular target \(Y\).
Continuity and Correct Endpoint Maps
A formula for \(H(x,s)\) defines a homotopy only if it is continuous as a function of both variables together. In particular, for a piecewise formula, checking continuity separately inside each piece is not enough: the pieces must also fit continuously along their boundary. The endpoint test is different and usually simpler: substitute \(s=0\) and \(s=1\), then compare the resulting functions with the maps claimed.
Worked Example: A Jump at the Parameter Boundary
Let \(X=Y=I\) with their usual subspace topologies. Consider the proposed formula \(H(x,s)=x\) for \(s<1/2\), and \(H(x,s)=1-x\) for \(s\geq1/2\). It is claimed to give a homotopy from the identity map to the reflection \(r(x)=1-x\).
For \(s=0\), the first case applies and \(H(x,0)=x\), as required. For \(s=1\), the second case applies and \(H(x,1)=1-x\), also as required. Both pieces take values in \(I\). But the formula is not continuous on \(I\times I\). At \(x=0\), the value is \(0\) whenever \(s<1/2\), while \(H(0,1/2)=1\). For example, the points \((0,1/2-1/n)\) approach \((0,1/2)\) as integers \(n\) tend to infinity, but their image is always \(0\), not \(1\). Thus continuity fails.
A valid replacement is \(K(x,s)=(1-s)x+s(1-x)\). It is continuous because it is a polynomial expression in the real coordinates \(x,s\). Its endpoints are \(K(x,0)=x\) and \(K(x,1)=1-x\). For \(x,s\in I\), both \(x\) and \(1-x\) belong to \(I\), and \(K(x,s)\) is their weighted average with nonnegative weights \(1-s,s\) summing to \(1\); hence \(K(x,s)\in I\).
Check: at \(x=0,s=1/2\), the corrected value is \((1-1/2)0+(1/2)(1-0)=1/2\), in \(I\). At \(s=1/2\), the formula varies continuously with \(x\), rather than switching between unrelated values. The corrected formula has the claimed endpoint maps and stays in the target.
Worked Example: The Formula Ends at the Wrong Map
Again take \(X=Y=I\). Suppose the goal is a homotopy from the identity map to the constant-zero map \(c_0(x)=0\), and the proposed formula is \(H(x,s)=(1-s)x+s/2\).
This expression is continuous. It takes values in \(I\): it is a weighted average of \(x\in I\) and \(1/2\in I\). At \(s=0\), it gives \(H(x,0)=x\). But at \(s=1\), it gives \(H(x,1)=1/2\), not \(0\). For a direct numerical check, set \(x=2/5\) and \(s=1\). Then \(H(2/5,1)=(1-1)(2/5)+1/2=1/2\), whereas \(c_0(2/5)=0\). The formula is a homotopy to the constant-\(1/2\) map, not to \(c_0\).
The corrected formula \(K(x,s)=(1-s)x\) is continuous, takes values in \(I\), and has \(K(x,0)=x\) and \(K(x,1)=0\). For example, \(K(2/5,1/4)=(3/4)(2/5)=3/10\), which lies in \(I\). Substitution at the two parameter endpoints verifies the intended maps exactly.
Keeping Path Endpoints Fixed
When the problem requires a homotopy with endpoints fixed, the endpoint conditions must hold for every value of the homotopy parameter. It is not sufficient for the initial and final paths to share endpoints. In the language of homotopy relative to a subspace, the fixed subset here is \(\{0,1\}\subseteq I\).
Worked Example: Correct Paths, Moving Endpoints
Let \(X=I\) and \(Y=\mathbb{R}\), both with their standard topologies or the subspace topology as appropriate. The path \(\alpha(t)=t\) has endpoints \(0\) and \(1\). Consider \(H(t,s)=t+s(1-s)\). This formula is continuous, and its two endpoint paths are both \(\alpha\): \(H(t,0)=t\) and \(H(t,1)=t\).
Nevertheless, it is not a homotopy with endpoints fixed. At the initial endpoint \(t=0\), the value is \(H(0,s)=s(1-s)\). Taking \(s=1/2\) gives \(H(0,1/2)=(1/2)(1/2)=1/4\), not \(0\). At \(t=1\), the value is \(H(1,s)=1+s(1-s)\), which at \(s=1/2\) is \(1+1/4=5/4\), not \(1\). The endpoint paths return to their required values at \(s=0\) and \(s=1\), but move in between.
A corrected homotopy relative to \(\{0,1\}\) is \(K(t,s)=t\). It is continuous, has \(K(t,0)=K(t,1)=\alpha(t)\), and satisfies \(K(0,s)=0\) and \(K(1,s)=1\) for every \(s\in I\).
Check: at \(t=0,s=1/2\), \(K(0,1/2)=0\); at \(t=1,s=1/2\), \(K(1,1/2)=1\). Both fixed-endpoint conditions hold throughout.
The example uses the same initial and final path, so it isolates the intermediate-endpoint test. In a proposed homotopy between distinct paths, first verify that their respective endpoints agree; then verify that the formula keeps those values fixed for all intermediate parameters.
Staying Inside the Target Space
Continuity of a formula into a larger space does not establish continuity as a map into the stated target. Before considering continuity, check that every output actually belongs to that target. For instance, the circle \(S^1=\{z\in\mathbb{C}:|z|=1\}\) has the subspace topology inherited from the Euclidean topology on \(\mathbb{C}\); a complex number of modulus less than \(1\) is not a point of \(S^1\).
Worked Example: A Straight Line Leaves the Circle
Let \(X=Y=S^1\), and consider the proposed formula \(H(z,s)=(1-s)z+s\). It is intended to join the identity map \(z\mapsto z\) to the constant map \(z\mapsto1\). The endpoint substitutions work: \(H(z,0)=z\) and \(H(z,1)=1\). The expression is continuous as a map into \(\mathbb{C}\), but it does not always lie in \(S^1\).
Take \(z=-1\in S^1\) and \(s=1/2\in I\). Then \(H(-1,1/2)=(1/2)(-1)+(1/2)(1)=0\), and \(|0|=0\ne1\). Thus \(H(-1,1/2)\notin S^1\), so the formula is not a map \(S^1\times I\to S^1\). It cannot be a homotopy in the stated target.
Changing the target to the closed unit disk \(D^2=\{z\in\mathbb{C}:|z|\leq1\}\), with its subspace topology, makes this particular formula valid. Indeed, for \(z\in S^1\), the triangle inequality gives \(|(1-s)z+s|\leq(1-s)|z|+s|1|=(1-s)+s=1\). The formula is continuous, has the stated endpoint maps, and takes values in \(D^2\). This verifies a homotopy into the disk, not into the circle.
For comparison, \(K(z,s)=e^{i\pi s}z\) is a valid homotopy in \(S^1\) from the identity to the antipodal map \(z\mapsto-z\). The factor \(e^{i\pi s}\) has modulus \(1\), so \(|K(z,s)|=|z|=1\); the formula is continuous and \(K(z,0)=z\), \(K(z,1)=-z\). At \(z=-1,s=1/2\), it gives \(K(-1,1/2)=e^{i\pi/2}(-1)=i(-1)=-i\), whose modulus is \(1\). This direct check confirms that this output remains on the circle.
A Reliable Construction in Convex Targets
A useful situation where the straight-line formula does work is a convex target. A subset \(Y\subseteq\mathbb{R}^n\) is convex if \((1-s)y_0+sy_1\in Y\) for every \(y_0,y_1\in Y\) and every \(s\in I\). Here \(\mathbb{R}^n\) has its Euclidean topology and \(Y\) its subspace topology.
Proof. For each \((x,s)\in X\times I\), both \(f(x)\) and \(g(x)\) belong to \(Y\). Convexity therefore gives \((1-s)f(x)+sg(x)\in Y\), so the formula has the required target. The coordinate functions of \(H\) are sums and products of continuous real-valued functions on \(X\times I\): the functions \(f\circ\pi_1\), \(g\circ\pi_1\), and the parameter projection \((x,s)\mapsto s\) are continuous, and addition and scalar multiplication in \(\mathbb{R}^n\) are continuous. Thus \(H\) is continuous into \(\mathbb{R}^n\), and, since its range lies in \(Y\), it is continuous into the subspace \(Y\). Finally, \(H(x,0)=f(x)\) and \(H(x,1)=g(x)\). If \(a\in A\), then \(f(a)=g(a)\), so \(H(a,s)=(1-s)f(a)+sf(a)=f(a)\) for every \(s\in I\). Hence it fixes \(A\) pointwise. \(\square\)
The theorem depends on convexity of the target, not merely on continuity of the endpoint maps. A circle is not convex as a subset of \(\mathbb{C}\): the segment between \(1\) and \(-1\) contains \(0\), which is not on the circle. That is exactly why the straight-line construction in the preceding example can leave \(S^1\).
A Practical Audit and Common Pitfall
For a proposed homotopy, use the following checks in order:
- Target: for every allowed input and parameter, does the formula land in \(Y\)?
- Endpoints: do substitution at \(s=0\) and \(s=1\) give exactly the required maps?
- Fixed subset: if a subset must stay fixed, does the formula have the required value there for every \(s\), not just at \(s=0,1\)?
- Continuity: is the formula continuous jointly in the input and parameter, including at any piecewise boundaries?
Common pitfall. Checking only the two endpoint maps and saying “the formula is continuous” does not prove a homotopy. For the circle formula \(H(z,s)=(1-s)z+s\), the endpoint calculation is correct, but the explicit substitution \(z=-1,s=1/2\) gives \(H=0\notin S^1\). The correction is not to repeat the continuity claim: first establish that the formula takes values in the target. If the target is the circle, use a formula whose values have modulus \(1\), such as \(e^{i\pi s}z\) for the identity-to-antipodal homotopy verified above.
Check Your Understanding
For each proposed formula, identify which homotopy condition must be checked or fails.
- For \(X=I\) and \(Y=I\), test whether \(H(x,s)=x+s/2\) can be a homotopy from the identity to the identity.
- For paths in \(\mathbb{R}\), determine whether \(H(t,s)=t+s\,t(1-t)\) fixes the endpoints of the path \(t\mapsto t\) throughout.
- Explain why convexity of the target is relevant to the straight-line homotopy theorem.
- For \(z\in S^1\), check whether \((1-s)z-s\) always belongs to \(S^1\), or find a counterexample.
- Give the four checks in the order you would apply them to a piecewise formula claimed to be a homotopy relative to a subset.